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anonymous

  • one year ago

A 2.0 µC point charge travels a distance of 0.20 cm between two parallel charged plates from the negative end toward the positive end. The electric field has a strength of 500.0 N.C. What is the change in voltage? (µC=1.0*10^-6 C)

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  1. anonymous
    • one year ago
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    A. -100 V B. +1.0 V C. 0 V D. +0.20 mV

  2. Michele_Laino
    • one year ago
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    the subsequent relationship holds: \[\Delta V = Ed\] |dw:1433273196060:dw|

  3. anonymous
    • one year ago
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    ok! so what happens next? :/

  4. anonymous
    • one year ago
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    500.0 x 0.2

  5. anonymous
    • one year ago
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    =100

  6. anonymous
    • one year ago
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    +1.0 V is your answer! :)

  7. anonymous
    • one year ago
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    ohh okay!! thank you!!:D

  8. Michele_Laino
    • one year ago
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    no, it is: \[\Large \Delta V = Ed = 500 \times 2 \times {10^{ - 3}} = ...\] since the distance is 0.2 cm = 2*10^(-3) meters

  9. anonymous
    • one year ago
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    so +1.0V is the solution?

  10. Michele_Laino
    • one year ago
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    yes! since our charge is approaching to the positive plane, so its potential is increasing, namlely final potential is larger than initial potential so the difference between them is positive

  11. anonymous
    • one year ago
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    yay! thank you both!!

  12. Michele_Laino
    • one year ago
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    :)

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